Linux 如何在Debian下制作php cronjob
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How to make a php cronjob under Debian
提问by Tom
I'm trying to set a cronjob to run every 20 minutes. The file path is /srv/www/mysite.co.uk/public_html/PP/Make_Xml.php
我正在尝试将 cronjob 设置为每 20 分钟运行一次。文件路径为/srv/www/mysite.co.uk/public_html/PP/Make_Xml.php
but i need to transfer to it a var so basically to cron: /srv/www/mysite.co.uk/public_html/PP/Make_Xml.php?db=LONDON
但我需要向它转移一个 var 所以基本上到 cron：/srv/www/mysite.co.uk/public_html/PP/Make_Xml.php?db=LONDON
I tried to use "crontab -e" and set it even to every minute with:
我尝试使用“crontab -e”并将其设置为每分钟：
* * * * * /srv/www/mysite.co.uk/public_html/PP/Make_Xml.php?db=LONDON
it saved it to /tmp/crontab.something/crontab
它保存到 /tmp/crontab.something/crontab
And it doesn't seem to work. I'm new to linux please help.
它似乎不起作用。我是 linux 新手，请帮忙。
采纳答案by Pascal MARTIN
First of all, when calling a PHP script from the command line, you will not pass it parameters the way you did here.
首先，当从命令行调用 PHP 脚本时，您不会像这里那样传递参数。
You'll typically pass those like this :
你通常会通过这样的：
/srv/www/mysite.co.uk/public_html/PP/Make_Xml.php db=LONDON
And, from your PHP script, you will not get the data into $_GET, but into $_SERVER['argv']
而且，从您的 PHP 脚本中，您不会将数据放入$_GET，而是放入$_SERVER['argv']
For example, if I create a temp.phpscript that contains this :
例如，如果我创建一个temp.php包含以下内容的脚本：
<?php
var_dump($_SERVER['argv']);
Calling it this way :
这样称呼它：
php temp.php db=LONDON
will get me the following output :
会给我以下输出：
array(2) {
[0]=>
string(8) "temp.php"
[1]=>
string(9) "db=LONDON"
}
Then, note you should probably call the phpexecutable program, from your crontab, and not directly the PHP script -- unless you made it executable.
然后，请注意，您可能应该从 crontab调用php可执行程序，而不是直接调用PHP 脚本——除非您使其可执行。
Which probably means using something like this :
这可能意味着使用这样的东西：
* * * * * /usr/bin/php /srv/www/mysite.co.uk/public_html/PP/Make_Xml.php db=LONDON
Note : you may need to adapt the path to php.
注意：您可能需要将路径调整为php.
回答by BMitch
Call it with php:
用 php 调用它：
* * * * * /usr/bin/php /srv/www/mysite.co.uk/public_html/PP/Make_Xml.php?db=LONDON
For the db=LONDON part, you may need to just pass LONDON as a command line arg. See the following site for more details on that: http://www.php.net/manual/en/features.commandline.usage.php
对于 db=LONDON 部分，您可能只需要将 LONDON 作为命令行参数传递。有关更多详细信息，请参阅以下站点：http: //www.php.net/manual/en/features.commandline.usage.php
回答by RDL
See Pascals response re: variables.
请参阅 Pascals 响应：变量。
For every 20 minutes you want:
您想要的每 20 分钟：
*/20 * * * * /usr/bin/php /srv/www/mysite.co.uk/public_html/PP/Make_Xml.php db=LONDON
Which says to run the script whenever the minutes is divisible by 20.
这表示只要分钟可被 20 整除就运行脚本。
回答by AndrewNimmo
There are 2 additional options you could consider.
您可以考虑另外 2 个选项。
For a PHP script which you intend using from the command-line, you can use the PEAR Console_Getoptcommand-line option parser which offers similar functionality to getopt libraries for other languages.
对于您打算从命令行使用的 PHP 脚本，您可以使用 PEAR Console_Getopt命令行选项解析器，它提供与其他语言的 getopt 库类似的功能。
If your PHP script needs to run in a web context, you can use a cron job to make requests for the URL. A stackoverflow question and answercovers this.
如果您的 PHP 脚本需要在 Web 上下文中运行，您可以使用 cron 作业来请求 URL。一个计算器问题和答案涵盖这一点。
